23. As shown in the figure, the side length of the square ABCD is 5cm. Use a triangular plate to make its right-angle edge always pass through point A, the right-angle vertex E moves on BC, and the other right-angle edge intersects CD at point F, if be = x cm, cf = y cm.

Try to use the algebraic expression of x to represent y (you don't need to write the range of x).

Solution: Triangle Abe is similar to triangle ECF.

AB=5，BE=X EC=5-X，CF = Y；

So AB/EC=BE/CF, that is, 5/(5-X)=X/Y,

You get Y=X minus 1/5 times the square of x.